SOLUTION: {{{ log3(n^2-5)^4=4 }}} I think this is how you start log base3(n^2-5)^4=3^4 (n^2-5)^4=81

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: {{{ log3(n^2-5)^4=4 }}} I think this is how you start log base3(n^2-5)^4=3^4 (n^2-5)^4=81      Log On


   

Question 281204: +log3%28n%5E2-5%29%5E4=4+
I think this is how you start
log base3(n^2-5)^4=3^4
(n^2-5)^4=81
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
>I think this is how you start
>log base3(n^2-5)^4=3^4

Actually the logarithm should be gone when you rewrite the equation in exponential form:
%28n%5E2-5%29%5E4=3%5E4

At this point, one way to solve the equation is to find the 4th root of each side:
root%284%2C+%28n%5E2-5%29%5E4%29+=+root%284%2C+3%5E4%29
Since 4th roots can be positive or negative we must account for both:
n%5E2+-+5+=+3 or n%5E2+-+5+=+-3
These are quadratic equations so we want one side equal to zero:
n%5E2+-+8+=+0 or n%5E2+-+2+=+0
These will not factor. So we'll use the Quadratic equation on each. We should end up with:
n+=+2sqrt%282%29 or n+=+-2sqrt%282%29 or n+=+sqrt%282%29 or n+=+-sqrt%282%29

When solving logarithmic equations you should check your answers. You must make sure no arguments of any logarithms are zero or negative.
+log%283%2C+%28n%5E2-5%29%5E4%29%29+=+4+
Checking n+=+2sqrt%282%29:
+log%283%2C+%28%282sqrt%282%29%29%5E2-5%29%5E4%29%29+=+4+
+log%283%2C+%28%288-5%29%5E4%29%29+=+4+
+log%283%2C+%283%5E4%29%29+=+4+ Check!

Checking n+=+2sqrt%282%29:
+log%283%2C+%28%28-2sqrt%282%29%29%5E2-5%29%5E4%29%29+=+4+
+log%283%2C+%28%288-5%29%5E4%29%29+=+4+
+log%283%2C+%283%5E4%29%29+=+4+ Check!

Checking n+=+sqrt%282%29:
+log%283%2C+%28%28sqrt%282%29%29%5E2-5%29%5E4%29%29+=+4+
+log%283%2C+%28%282-5%29%5E4%29%29+=+4+
+log%283%2C+%28%28-3%29%5E4%29%29+=+4+ Check!

Checking n+=+-sqrt%282%29:
+log%283%2C+%28%28-sqrt%282%29%29%5E2-5%29%5E4%29%29+=+4+
+log%283%2C+%28%282-5%29%5E4%29%29+=+4+
+log%283%2C+%28%28-3%29%5E4%29%29+=+4+ Check!